Concept:

For ‘x’ displacement of mass, spring deflection is given by

x_s = x cos α

∴ F_s = k x_s = kx cos α 

∴ k_eq x = [k (cos α) x] cos α

∴ k_eq = k cos² α

Calculation:

k_eq = k cos² α

∴ k_eq = 0.75 k 

Concept:

The critical system is when the damping ratio become 1.

Natural frequency:

${\omega }_n=\sqrt{\frac{k}{m}}$

2ξω_n = c/m

Damping ratio:

$\xi =\frac{c}{c_c}=\frac{c}{2\sqrt{km}}=\frac{c}{2m{\omega }_n}$

Calculation:

m = 15.7 kg, k = 1000 N/m

For ξ = 1

$c=c_c=2\sqrt{km}=2\sqrt{1000\times 15.7}=250.60rad/s$

Concept:

Damping ratio:

$\xi =\frac{C}{C_c}$

ξ > 1: Overdamped system

ξ = 1: Critically damped system

ξ < 1: Under-damped system

Differential equations of damped free vibration

$m\ddot{x}+c\dot{x}+kx=0$

Calculation:

2ẍ + 4ẋ + 16x = 0

ẍ + 2ẋ + 8x = 0

On comparing with: mẍ + cẋ + kx = 0

M = 1 kg, c = 2 Ns/m, k = 8 N/m

$\xi =\frac{c}{2\sqrt{km}}=\frac{2}{2\sqrt{8\times 1}}=0.353$

ξ < 1 ⇒ underdamped system

Explanation:

Given:

Mass (m) = 25 kg, k = 22,000 N/m, Damping provided = 45 %

$\xi =\frac{C}{C_c}=0.45$

$\mathrm{N}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\left({\omega }_n\right)=\sqrt{\frac{k}{m}}$

${\omega }_n=\sqrt{\frac{22000}{25}}$

∴ ω_n = 29.66 rad/s

Damped frequency (ω_d)

${\omega }_d={\omega }_n\sqrt{1-{\xi }^2}$

∴ ω_d = 26.49 rad/s

Time period = 2π/ω

∴ T = 0.237 sec 

m = 7 kg, k = 3 kN/m = 3000 N/m, c = 150 Ns/m, x₁ = 13 mm

Concept:

The critical system is when the damping ratio become 1.

Natural frequency:

${\omega }_n=\sqrt{\frac{k}{m}}$

2ξω_n = c/m

Damping ratio:

$\xi =\frac{c}{c_c}=\frac{c}{2\sqrt{km}}=\frac{c}{2m{\omega }_n}$

Logarithmic Decrement:

$\delta =\ln \frac{x_0}{x_1}=\ln \frac{x_1}{x_2}$

If the system executes n cycles:

$\delta =\frac{1}{n}\ln \frac{x_o}{x_n}=\frac{2\pi \xi }{\sqrt{1-{\xi }^2}}$

$\xi =\frac{\delta }{\sqrt{4{\pi }^2+{\delta }^2}}$

Calculation:

${\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{3000}{7}}=20.7rad/s$

$\xi =\frac{c}{2\sqrt{km}}=\frac{150}{2\sqrt{3000\times 7}}=0.52$

X₁ is the amplitude at first cycle and X₂ at the second cycle.

T is the time period of the damped vibration

Logarithmic decrement:

$\delta =\frac{2\pi \xi }{\sqrt{1-{\xi }^2}}=\frac{2\pi \times 0.52}{\sqrt{1-{\left(0.52\right)}^2}}=3.825$

$\delta =\ln \frac{x_1}{x_2}\Rightarrow \frac{x_1}{x_2}=e^{\delta }=e^{3.825}=45.833$

$x_2=\frac{x_1}{45.833}=\frac{13}{45.833}=0.284$

Explanation:

First, we need to find the stiffness of rod,

Deflection is given as

$\delta =\frac{W{L}^3}{3EI}=\frac{mg{L}^3}{3EI}$

$\therefore k_1=\frac{mg}{\delta }=\frac{3EI}{L^3}$

Given: m = 8 kg, L = 1 m, E = 210 GPa, A = (20 × 20) mm², k = 5 kN/m

Now, stiffness of rod and stiffness of spring forms a series combination.

$k_{eq}=\frac{k_1\times k}{k_1+k}$

$k_{eq}=\frac{\frac{3EI}{L^3}\times k}{\frac{3EI}{L^3}+k}$

$I=\frac{20\times {20}^3}{12}=13333.33m{m}^4=1.3333\times {10}^{-8}{m}^4$

E = 210 × 10⁹ Pa

$k_1=\frac{3EI}{L^3}=\frac{3\times 210\times {10}^9\times 1.333\times {10}^{-8}}{{\left(1\right)}^3}$

k₁ = 8400 N/m

$\therefore k_{eq}=\frac{8400\times 5000}{8400+5000}$

∴ k_eq = 3134.328 N/m

Now,

${\omega }_n=\sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{3134.28}{8}}$

∴ ω_n = 19.79 rad/s ≈ 19.8 rad/s 

Concept:

Natural frequency:

${\omega }_n=\sqrt{\frac{k}{m}}$

Damping factor:

$\xi =\frac{c}{c_c}=\frac{c}{2\sqrt{km}}=\frac{c}{2m{\omega }_n}$

Amplitude of vibration:

$A=\frac{\frac{F_o}{k}}{\sqrt{{\left(1-{\left(\frac{\omega }{{\omega }_n}\right)}^2\right)}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}=\frac{X_s}{\sqrt{{\left(1-{\left(\frac{\omega }{{\omega }_n}\right)}^2\right)}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}$

Calculation:

Given: m = 15 kg, k = 3375 N/m, F = F_o sin(wt), F_o = 23 N, ω = 15 rad/sec, A = 35 mm = 35 × 10^-3 m

The natural undamped frequency be (ω_n)

${\omega }_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{3375}{15}}=15rad/s$

Here: ω = ω_n = 15 rad/sec

ω/ω_n = 1

$\therefore A=\frac{\frac{F_o}{k}}{2\xi }=\frac{F_o}{2\xi k}$

$35\times {10}^{-3}=\frac{23}{2\xi \times 3375}\Rightarrow \xi =0.097$

$\xi =\frac{c}{2m{\omega }_n}$

$\Rightarrow c=2m{\omega }_n\xi =2\times 15\times 15\times 0.097=43.65Ns/m$

Explanation:

Let the deflection of spring be ‘x’

$\mathrm{D}\mathrm{e}\mathrm{f}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}=\frac{Lx}{a}$

The dampers are in series

$\therefore {\mathrm{c}}_{\mathrm{E}\mathrm{q}}=\frac{c\times 2c}{c+2c}$

$\therefore {\mathrm{c}}_{\mathrm{E}\mathrm{q}}=\frac{2}{3}c$

Now, the Equivalent equation.

$m\left(\frac{L^2}{a}\ddot{x}\right)+\frac{2}{3}c\dot{x}a+3kxa=0$

$\mathrm{D}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}=\frac{2c{a}^2/3L^2}{2\sqrt{\frac{m{a}^2}{L^2}3k}}$

∴ Damping ratio = 1

Now,

$\frac{c{a}^2}{3L^2}=\frac{a}{L}\sqrt{3mk}$

$c=\frac{3L}{a}\sqrt{3mk}$

Concept:

$\delta =ln\left(\frac{x_1}{x_2}\right)$

$\xi =\frac{\delta }{\sqrt{4{\pi }^2+{\delta }^2}}$

${\omega }_d={\omega }_n\sqrt{1-{\xi }^2}$

$C=2\xi \sqrt{km}$

Calculation:

m = 1 kg        T_d = 0.2 sec

$\delta ln\left[\frac{0.9}{0.4}\right]=0.811$

$\xi =\frac{\delta }{\sqrt{4{\pi }^2+{\delta }^2}}=0.128$

${\omega }_d=\frac{2\pi }{T_d}=\frac{2\pi }{0.2}=31.416rad/s$

${\omega }_d={\omega }_n\sqrt{1-{\xi }^2}$

${\omega }_n=\frac{31.416}{\sqrt{1-{0.128}^2}}=31.67rad/s$

${\omega }_n=\sqrt{\frac{k}{m}}\Rightarrow k=m{\omega }_n^2$

∴ k = 1 × 31.67² = 1003.4 N/m

Now,

$C=2\xi \sqrt{km}$

$C=2\times 0.128\times \sqrt{1003.4\times 1}$

∴ C = 8.1 Ns/m

Concept:

Motion of Forced damped vibration:

$m\ddot{x}+c\dot{x}+kx=F\cos \left(\omega t\right)$ ------(1)

${\omega }_n=\sqrt{\frac{K_{eq}}{m}}$

Damping factor: $\xi =\frac{c}{c_c}=\frac{c}{2\sqrt{km}}$

$T=\frac{F_T}{F_{un}}=\frac{\sqrt{1+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}{\sqrt{{(1-{\left(\frac{\omega }{{\omega }_n}\right)}^2)}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}$

Calculation:

Given:

$3\ddot{x}+14\dot{x}+72x=200\cos \left(9t\right)$

On comparing the above equation with equation 1

m = 3 kg, c = 14 Ns/m, K = 72 N/m, ω = 9 rad/s, F = 200 N

${\omega }_n=\sqrt{\frac{K_{eq}}{m}}=\sqrt{\frac{72}{3}}=4.9rad/s$

$\xi =\frac{c}{2\sqrt{km}}=\frac{14}{2\times \sqrt{72\times 3}}=0.4763$

$\frac{\omega }{{\omega }_n}=\frac{9}{4.9}=1.837$

$T=\frac{F_T}{F_{un}}=\frac{\sqrt{1+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}{\sqrt{{(1-{\left(\frac{\omega }{{\omega }_n}\right)}^2)}^2+{\left(2\xi \frac{\omega }{{\omega }_n}\right)}^2}}$

$T=\frac{\sqrt{1+{(2\times 0.4763\times 1.837)}^2}}{\sqrt{{(1-{\left(1.837\right)}^2)}^2+{(2\times 0.4763\times 1.837)}^2}}=\frac{2.015}{2.95}=0.683$

$F_T=T\times F_{un}=0.683\times 200=136.6N$